3.7.41 \(\int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx\) [641]

3.7.41.1 Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\frac {a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}+\frac {d \left (b^2-a^2 (1+m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right ) (d \sec (e+f x))^{-1+m} \sin (e+f x)}{f (1-m) (1+m) \sqrt {\sin ^2(e+f x)}}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)} \]

output

a*b*(2+m)*(d*sec(f*x+e))^m/f/m/(1+m)+d*(b^2-a^2*(1+m))*hypergeom([1/2, -1/ 
2*m+1/2],[3/2-1/2*m],cos(f*x+e)^2)*(d*sec(f*x+e))^(-1+m)*sin(f*x+e)/f/(-m^ 
2+1)/(sin(f*x+e)^2)^(1/2)+b*(d*sec(f*x+e))^m*(a+b*tan(f*x+e))/f/(1+m)
 

3.7.41.2 Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.81 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\frac {(d \sec (e+f x))^m \left (b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right ) \tan (e+f x)+a \left (-a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right ) \tan (e+f x)+2 b \sqrt {-\tan ^2(e+f x)}\right )\right )}{f m \sqrt {-\tan ^2(e+f x)}} \]

input

Integrate[(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^2,x]
 

output

((d*Sec[e + f*x])^m*(b^2*Hypergeometric2F1[-1/2, m/2, (2 + m)/2, Sec[e + f 
*x]^2]*Tan[e + f*x] + a*(-(a*Hypergeometric2F1[1/2, m/2, (2 + m)/2, Sec[e 
+ f*x]^2]*Tan[e + f*x]) + 2*b*Sqrt[-Tan[e + f*x]^2])))/(f*m*Sqrt[-Tan[e + 
f*x]^2])
 

3.7.41.3 Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3993, 25, 3042, 3967, 3042, 4259, 3042, 3122}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (a+b \tan (e+f x))^2 (d \sec (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (e+f x))^2 (d \sec (e+f x))^mdx\)

\(\Big \downarrow \) 3993

\(\displaystyle \frac {\int -(d \sec (e+f x))^m \left (-\left ((m+1) a^2\right )-b (m+2) \tan (e+f x) a+b^2\right )dx}{m+1}+\frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)}-\frac {\int (d \sec (e+f x))^m \left (-\left ((m+1) a^2\right )-b (m+2) \tan (e+f x) a+b^2\right )dx}{m+1}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)}-\frac {\int (d \sec (e+f x))^m \left (-\left ((m+1) a^2\right )-b (m+2) \tan (e+f x) a+b^2\right )dx}{m+1}\)

\(\Big \downarrow \) 3967

\(\displaystyle \frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)}-\frac {\left (b^2-a^2 (m+1)\right ) \int (d \sec (e+f x))^mdx-\frac {a b (m+2) (d \sec (e+f x))^m}{f m}}{m+1}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)}-\frac {\left (b^2-a^2 (m+1)\right ) \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx-\frac {a b (m+2) (d \sec (e+f x))^m}{f m}}{m+1}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)}-\frac {\left (b^2-a^2 (m+1)\right ) \left (\frac {\cos (e+f x)}{d}\right )^m (d \sec (e+f x))^m \int \left (\frac {\cos (e+f x)}{d}\right )^{-m}dx-\frac {a b (m+2) (d \sec (e+f x))^m}{f m}}{m+1}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)}-\frac {\left (b^2-a^2 (m+1)\right ) \left (\frac {\cos (e+f x)}{d}\right )^m (d \sec (e+f x))^m \int \left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}\right )^{-m}dx-\frac {a b (m+2) (d \sec (e+f x))^m}{f m}}{m+1}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)}-\frac {-\frac {d \left (b^2-a^2 (m+1)\right ) \sin (e+f x) (d \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{f (1-m) \sqrt {\sin ^2(e+f x)}}-\frac {a b (m+2) (d \sec (e+f x))^m}{f m}}{m+1}\)

input

Int[(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^2,x]
 

output

-((-((a*b*(2 + m)*(d*Sec[e + f*x])^m)/(f*m)) - (d*(b^2 - a^2*(1 + m))*Hype 
rgeometric2F1[1/2, (1 - m)/2, (3 - m)/2, Cos[e + f*x]^2]*(d*Sec[e + f*x])^ 
(-1 + m)*Sin[e + f*x])/(f*(1 - m)*Sqrt[Sin[e + f*x]^2]))/(1 + m)) + (b*(d* 
Sec[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(1 + m))
 

3.7.41.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3967

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a   Int[(d 
*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] 
|| NeQ[a^2 + b^2, 0])
 

rule 3993

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^2, x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m 
 + 1))), x] + Simp[1/(m + 1)   Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^2 + 
a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^ 
2 + b^2, 0] &&  !IntegerQ[m]
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

3.7.41.4 Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )\right )^{2}d x\]

input

int((d*sec(f*x+e))^m*(a+b*tan(f*x+e))^2,x)
 

output

int((d*sec(f*x+e))^m*(a+b*tan(f*x+e))^2,x)
 

3.7.41.5 Fricas [F]

\[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]

input

integrate((d*sec(f*x+e))^m*(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

output

integral((b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2)*(d*sec(f*x + e))^ 
m, x)
 

3.7.41.6 Sympy [F]

\[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{m} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \]

input

integrate((d*sec(f*x+e))**m*(a+b*tan(f*x+e))**2,x)
 

output

Integral((d*sec(e + f*x))**m*(a + b*tan(e + f*x))**2, x)
 

3.7.41.7 Maxima [F]

\[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]

input

integrate((d*sec(f*x+e))^m*(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

output

integrate((b*tan(f*x + e) + a)^2*(d*sec(f*x + e))^m, x)
 

3.7.41.8 Giac [F]

\[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]

input

integrate((d*sec(f*x+e))^m*(a+b*tan(f*x+e))^2,x, algorithm="giac")
 

output

integrate((b*tan(f*x + e) + a)^2*(d*sec(f*x + e))^m, x)
 

3.7.41.9 Mupad [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]

input

int((d/cos(e + f*x))^m*(a + b*tan(e + f*x))^2,x)
 

output

int((d/cos(e + f*x))^m*(a + b*tan(e + f*x))^2, x)